Ledge Problems • 2 Types (Both Solved with Torque) – Board held by a cable 55.0 0 – Traditional Ladder We can solve this problem by adding torques. We have three forces acting on the beam. 1. The weight of the beam, acts at the beams center (its CG), FB. 2. The tension in the cable, T. 3. And the force exerted by the wall on the beam, R. (The wall is pushing the beam up and out.) Pivot point: where the beam meets the wall. Therefore R exerts no torque as its lever arm is zero. We only have two torques, they add up to zero. Torque #1 is exerted by the tension in the cable Torque #2 is caused by the weight of the beam. Only the vertical component of the tension causes torque : cable B 0 r Tr sin FB 0 2 T cos RX 0 TO FIND T ON CABLE cable B 0 Tr sin FB rB 0 Tr sin FB rB TO FIND FORCE (R) OF WALL Rx: Ry: T cos RX 0 RY FB T sin TENSION ON CABLE The 2.4 m. long weightless beam shown in the figure is supported on the right by a cable that makes an angle of 50o with the horizontal beam. A 32 kg mass hangs from the beam 1.5 m from the pivot point on the left. Find the T on cable. Tr sin FM rM 0 FM rM T r sin x m = 32 kg (32kg )(9.8)(1.5m) T 2.4 sin 50 470N 256 N T 1.84 RESULTANT X FORCE (Rx)OF WALL RX T cos RESULTANT Y FORCE (Ry)OF WALL RESULTANT Y FORCE (R)OF WALL Ry Rx R ( Rx ) 2 ( R y ) 2 The 2.4 m. long weightless beam shown in the figure is supported on the right by a cable that makes an angle of 50o with the horizontal beam. A 32 kg mass hangs from the beam 1.5 m from the pivot point on the left. Find the R of wall. T 256 N b) y direction: T sin R y 0 R y (256 N ) sin 50.0o x m = 32 kg R y 196.1N x direction: T cos Rx 0 (256 N ) cos 50.0 Rx 0 o R y 314 N 196.1N 118 N R ( Rx ) 2 ( R y ) 2 Rx 164.6 N R (164.6 N ) 2 (118 N ) 2 140.0 N A uniform beam is supported by a stout piece of line as shown. The beam weighs 175 N. The cable makes an angle of 75.0. Find (a) the tension in the cable and (b) the force exerted on the end of the beam by the wall. Tr sin FB rB 0 T FB r 2 r sin 175 N 2 sin 75.0o 75.0 4.00 m 90.6 N (b) Next we can sum up the forces: x direction: T cos RX 0 T RX T cos y direction: 75.0 90.6 N cos 75.0 23.4 N FB RY FB T sin 175 N 90.6 N sin 75.0o R Ry 2 Rx 2 R 87.5 N 2 23.4 N 2 87.5 N 90.6 N A uniform beam of weight 254 N sticks out from a vertical wall. A lightweight cable connects the end of the beam to the wall, making an angle of 65.0 between the beam and the cable. (a) What is the tension in the cable? (b) What is the force exerted on the beam by the wall? a) Tr sin FB rB 0 254 N T 140.1N o 2 sin 65.0 FBeam T 2sin b) y direction: T sin R y FB 0 R y (254 N ) (140 N ) cos 65.0o R y 127.1N x direction: T cos Rx 0 R Ry Rx 2 2 (140 N ) cos 65.0o Rx 0 Rx 59.17 N 127.1 N 59.17 N 140.0 N 2 2 Tr sin FB rB Find Rx T cos Rx 0 Find Ry R y FB T sin Fm rm FB rB Tr sin 0 R x T cos Tr sin Fm rm FB rB 0 TO FIND T ON CABLE Tr sin Fm rm FB rB 0 TO FIND F ON WALL R y FM FB T sin R x T cos R ( Rx ) 2 ( R y ) 2 The 2.4 m. long beam (30kg) shown in the figure is supported on the right by a cable that makes an angle of 50o with the horizontal beam. A 32 kg mass hangs from the beam 1.5 m from the pivot point on the left. A) Determine the cable tension needed to produce equilibrium. b) the force exerted on the end of the beam by the wall. Fm rm FB rB Tr sin 0 470 N 352 Fm rm FB rB T 448 N T o sin 50.0 (2.4) O sin r b) x Find Rx m = 32 kg R T cos X T cos Rx 0 o R ( 448 N ) cos 50 . 0 288 N x Find Ry T sin R y FB Fm 0 R y FB Fm T sin a) R y (294 N ) (313N ) (448 N ) sin 50.0 R y 264 N o R ( Rx ) 2 ( R y ) 2 R ( 288) 2 ( 264) 2 A beam is supported as shown. The beam is uniform and weighs 300.0 N and 5.0 m long. A 635 N person stands 1.50 m from the building. (a) What is the tension in the cable? (b) the force exerted on the end of the beam by the wall. (a) Sum of torques: beam man cable 0 Fm rm FB rB Tr sin 0 Fm rm FB rB T r sin (635 N )(1.5m) (300 N )(2.5m) T (5m) sin 55o 55.0 0 R T 55.0 FB T 416 N Fm (b) the force exerted on the end of the beam by the wall. Find Rx RX T cos o Rx (416 N ) cos 55.0 238 N T cos Rx 0 Find Ry Ry FB Fm T sin T sin R y FB Fm 0 o R y (300 N ) (635 N ) (416 N ) sin 55.0 R y 624 N R R y Rx 2 2 624 N 2 238 N 2 668 N
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